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The limit of both the numerator and the denominator is $0$, but how to find the limit of the fraction $\lim_{x\to 1}\frac{1-\sqrt[n]{\cos{2n\pi x}}}{(x-1)(x^x-1)}$

I tried to use L 'Hospital's rule, but the limit of the numerator and denominator is still $0$.

And the cosine function is in the radical, and I don't think the Taylor expansion of trigonometric functions can do that either.

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    $\begingroup$ We know $e^x-1 \sim x, x\to 0$ and You can use it twice: in nominator and denominator. $\endgroup$
    – zkutch
    Commented Oct 19, 2023 at 13:55

2 Answers 2

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Some hint steps: as I wrote in comment, let's change $1-\sqrt[n]{\cos{2n\pi x}}=-(e^{\frac{1}{n}\cdot\ln\cos(2n\pi x)}-1)\sim -\frac{1}{n}\ln\cos(2n\pi x)$ when $x\to 1$.

In same way $x^x-1=e^{x\ln x}-1\sim x\ln x \sim \ln x \sim (x-1), x\to 1$.

Now you can use L 'Hospital and for last step remember, that $\sin (2\pi nx)=\sin (2\pi nx-2\pi n)$. Hope, you can collect all and come to answer.

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  • $\begingroup$ cool! maybe the result is $2n\pi^2$ $\endgroup$
    – S.Y.Li
    Commented Oct 19, 2023 at 14:39
  • $\begingroup$ I might agree with yours "maybe", if I knew what you were putting into it. $\endgroup$
    – zkutch
    Commented Oct 20, 2023 at 1:39
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Set $x=1+t$ and note that $$\cos2\pi n (1+t) = \cos 2\pi n t$$

Now, you can use the following known (and easily to verify) results $$\lim_{v \to 1}\frac{1-\sqrt[n]{v}}{1-v} = \frac 1n,\:\:\lim_{u \to 0}\frac{1-\cos u}{u^2} = \frac 12,\:\: \frac d{dt}\left[(1+t)^{1+t}\right]_{t=0} = 1$$ by "unnesting" the expression as follows: \begin{eqnarray*} \frac{1-\sqrt[n]{\cos{2n\pi x}}}{(x-1)(x^x-1)} & \stackrel{x=1+t}{=} & \frac{1-\sqrt[n]{\cos 2\pi n t}}{1-\cos 2\pi n t}\cdot \frac{1-\cos 2\pi n t}{(2\pi n t)^2}\cdot(2\pi n)^2\cdot \frac t{(1+t)^{1+t}-1} \\ & \stackrel{t\to 0}{\rightarrow} & \frac 1n \cdot \frac 12 \cdot (2\pi n)^2 \cdot 1 \\ & = & 2\pi^2 n \end{eqnarray*}

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